Inverter system, flyback dc

Inverter system, flyback dc
Lecture 6
ECEN 4517/5517
Experiment 4: inverter system
12 VDC
Battery
HVDC: 120 - 200 VDC
DC-DC
converter
DC-AC
inverter
+
Isolated
flyback
H-bridge
vac(t)
AC load
120 Vrms
60 Hz
–
d(t)
Feedback
controller
d(t)
Vref
–+
Step-up dc-dc converter
with isolation (flyback)
DC-AC inverter (H-bridge)
Feedback controller to
regulate HVDC
ECEN 4517
Digital
controller
1
Due dates
Right now:
Prelab assignment for Exp. 3 Part 3 (one from every student)
Due within five minutes of beginning of lecture
This week in lab (Feb. 19-21):
Nothing due. Try to finish Exp. 3.
Next week in lecture (Feb. 26):
Prelab assignment for Exp. 4 Part 1 (one from every student)
Next week in lab (Feb. 26-28):
Definitely finish Exp. 3, and begin Exp. 4
The following week in lab (Mar. 4-6):
Exp. 3 final report due
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2
Goals in upcoming weeks
Exp. 4: A three-week experiment
vHVDC
Exp. 4 Part 1:
Design and fabrication of
flyback transformer
Snubber circuit
Demonstrate flyback
converter power stage
operating open loop
Vbatt
snubber
PWM
Compensator
Exp. 4 Part 2:
Design feedback loop
Measure loop gain, compare with simulation and theory
Demonstrate closed-loop control of converter output voltage
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3
–+
Vref
Exp. 4, Part 3
H-bridge inverter, off grid
vHVDC
IR3101
IR3101
iac(t)
+
• Filtering of ac output not explicitly shown
AC load
120 Vrms
60 Hz
vac(t)
–
Digital
controller
• IR 3101 half-bridge modules with
integrated drivers
Exp. 4 Part 3: off-grid inverter
• Grid-tied: control iac(t)
• Demonstrate modified sine-wave inverter
(required)
• Off-grid: control vac(t)
• Demonstrate PWM inverter (extra credit)
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4
“Modified Sine-Wave” Inverter
vac(t)
vac(t) has a
rectangular
waveform
Inverter transistors
switch at 60 Hz,
T = 8.33 msec
+ VHVDC
DT/2
T/2
– VHVDC
• Choose VHVDC larger than
desired Vac,RMS
RMS value of vac(t) is:
Vac,RMS =
1
T
T
2
vac t dt = D V HVDC
0
• Can regulate value of
Vac,RMS by variation of D
• Waveform is highly
nonsinusoidal, with
significant harmonics
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5
PWM Inverter
Average vac(t) has a
sinusoidal waveform
Inverter transistors
switch at frequency
substantially higher
than 60 Hz
vac(t)
t
• Choose VHVDC larger than desired Vac,peak
• Can regulate waveshape and value of Vac,RMS by variation of d(t)
• Can achieve sinusoidal waveform, with negligible harmonics
• Higher switching frequency leads to more switching loss and
need to filter high-frequency switching harmonics and commonmode currents
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6
The buck-boost converter
1
Vg
2
iL
+
–
+
Switch in position 1: Vg charges
inductor
V
Switch in position 2: energy stored in
inductor is transferred to output
–
Conversion ratio:
V =– D
Vg
1–D
Subinterval 1
iL
Vg
+
–
Subinterval 2
+
iL
Vg
V
V
–
–
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+
–
+
7
See also:
supplementary
notes on Flyback
converter, Exp. 4
web page
The flyback converter:
A transformer-isolated buck-boost converter
Q1
buck-boost converter:
D1
–
Vg
+
–
L
V
+
Q1
construct inductor
winding using two
parallel wires:
D1
1:1
Vg
+
–
L
–
V
+
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8
Derivation of flyback converter, cont.
Isolate inductor
windings: the flyback
converter
Q1
D1
–
1:1
Vg
+
–
V
LM
+
Flyback converter
having a 1:n turns
ratio and positive
output:
1:n
Vg
+
–
LM
Q1
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9
+
D1
C
V
–
A simple transformer model
Multiple winding transformer
i1(t)
n1 : n2
Equivalent circuit model
i1(t)
i2(t)
+
+
+
v1(t)
v2(t)
v1(t)
–
–
i1'(t)
n1 : n 2
+
iM(t)
LM
v2(t)
–
–
i3(t)
i3(t)
+
v3(t)
+
v1(t) v2(t) v3(t)
n 1 = n 2 = n 3 = ...
0 = n 1i 1' (t) + n 2i 2(t) + n 3i 3(t) + ...
v3(t)
–
–
: n3
: n3
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i2(t)
Ideal
transformer
10
The magnetizing inductance LM
Transformer core B-H characteristic
• Models magnetization of
transformer core material
B(t) ∝
• Appears effectively in parallel with
windings
• If all secondary windings are
disconnected, then primary winding
behaves as an inductor, equal to the
magnetizing inductance
saturation
slope ∝ LM
H(t) ∝ i M (t)
• At dc: magnetizing inductance tends
to short-circuit. Transformers cannot
pass dc voltages
• Transformer saturates when
magnetizing current iM is too large
ECEN 4517
v1(t) dt
11
Volt-second balance in LM
The magnetizing inductance is a real inductor,
obeying
i1(t)
di (t)
v1(t) = L M M
dt
+
v1(t)
integrate:
i M (t) – i M (0) = 1
LM
t
0
ECEN 4517
Ts
0
n1 : n 2
i2(t)
+
iM(t)
LM
v2(t)
–
v1(τ)dτ
Magnetizing current is determined by integral of
the applied winding voltage. The magnetizing
current and the winding currents are independent
quantities. Volt-second balance applies: in
steady-state, iM(Ts) = iM(0), and hence
0= 1
Ts
i1'(t)
–
i3(t)
+
v3(t)
–
: n3
Ideal
transformer
v1(t)dt
12
The “flyback transformer”
Transformer model
ig
+
–
Vg
i
+
LM
vL
1:n
+
iC
D1
C
R
v
–
–
Q1
A two-winding inductor
Symbol is same as
transformer, but function
differs significantly from
ideal transformer
Energy is stored in
magnetizing inductance
Magnetizing inductance is
relatively small
Current does not simultaneously flow in primary and secondary windings
Instantaneous winding voltages follow turns ratio
Instantaneous (and rms) winding currents do not follow turns ratio
Model as (small) magnetizing inductance in parallel with ideal transformer
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13
Subinterval 1
Transformer model
ig
Vg
+
–
i
+
LM
vL
+
iC
1:n
C
R
v
–
–
vL = V g
iC = – v
R
ig = i
CCM: small ripple
approximation leads to
Q1 on, D1 off
ECEN 4517
vL = V g
iC = – V
R
ig = I
14
Subinterval 2
Transformer model
ig
=0
Vg
+
–
+
vL
–
i
i/n
–
v/n
+
+
iC
1:n
C
R
v
–
vL = – nv
i C = ni – v
R
ig = 0
CCM: small ripple
approximation leads to
vL = – V
n
i C = nI – V
R
ig = 0
Q1 off, D1 on
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15
CCM Flyback waveforms and solution
vL
Vg
Volt-second balance:
vL = D Vg + D' – V
n =0
–V/n
Conversion ratio is
M(D) = V = n D
Vg
D'
Charge balance:
I/n – V/R
iC
i C = D – V + D' nI – V = 0
R
R
–V/R
Dc component of magnetizing
current is
ig
I
I = nV
D'R
Dc component of source current is
0
DTs
Conducting
devices:
ECEN 4517
Ts
Q1
D'Ts
I g = i g = D I + D' 0
t
D1
16
Equivalent circuit model: CCM Flyback
vL = D Vg + D' – V
n =0
i C = D – V + D' nI – V = 0
R
R
Vg
+
–
+
I
Ig
DI
+ DV
g
–
D'V
n
D'I
n
+
–
I g = i g = D I + D' 0
D' : n
+
I
Ig
+
–
R
V
–
ECEN 4517
V
–
1:D
Vg
R
17
Step-up DC-DC flyback converter
Need to step up the 12 V battery voltage to HVDC (120-200 V)
How much power can you get using the parts in your kit?
Key limitations:
MOSFET on-resistance (90 m)
Input capacitor rms current rating:
25 V 2200 μF: 2.88 A
35 V 2200 μF: 3.45 A
Snubber loss
Need to choose turns ratio, as well as D, fs, to minimize peak currents
Possible project for expo: build a better (and more complex) step-up dc-dc
converter
ECEN 4517
18
Design of CCM flyback transformer
iM(t)
n1 : n2
i1
Vg
+
–
iM
+
LM
vM
+
D1
C
R
V
0
i1(t)
IM
–
–
i2
Q1
0
i2(t)
n1
I
n2 M
0
vM(t)
0
ECEN 4517
ΔiM
IM
Transformer model
19
Vg
DTs
Approach
Use your PQ 32/20 core
Choose turns ratio n2/n1, LM, D, and fs (choose your own
values, dont use values in supplementary notes)
Select primary turns n1 so that total loss Ptot in flyback
transformer is minimized:
Ptot = Pfe + Pcu = core loss plus copper loss
Determine air gap length
Determine primary and secondary wire gauges
Make sure that core does not saturate
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20
Core loss
CCM flyback example
B-H loop for this application:
The relevant waveforms:
B(t)
B(t)
Bsat
Bmax
Bmax
ΔB
0
Hc(t)
vM(t)
Minor B–H loop,
CCM flyback
example
0
ΔB
Vg
n1 A c
Vg
DTs
B–H loop,
large excitation
B(t) vs. applied voltage,
from Faradays law:
dB(t) vM (t)
=
n1 A c
dt
Solve for B:
ECEN 4517
21
For the first
subinterval:
VgDTs
B=
2n 1A c
Vg
dB(t)
=
n1 A c
dt
Calculation of ac flux density
and core loss
Fitting an equation to the plot at right
P fe = K fe(ΔB) β A c l m
= slope
Kfe = constant that depends on fs
Aclm = core volume
At 60˚C:
= 2.6
Kfe = 16 (50 kHz), 40 (100 kHz)
with Pfe in watts, Aclm in cm3, B in Tesla
From previous slide:
VgDTs
B=
2n 1A c
More turns less B less core loss
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22
Copper loss
Power loss in resistance of wire
Must allocate the
core window area
between the various
windings
Winding 1 allocation
α1WA
Winding 2 allocation
α2WA
0 < αj < 1
α1 + α2 +
{
{
Total window
area WA
etc.
Optimum
choice:
αm =
n mI m
∞
Σ nI
n=1
(leads to minimum total copper loss)
j j
ρ(MLT )n 21 I 2tot
The resulting total copper loss is: Pcu =
WAK u
Choose wire gauges:
ECEN 4517
+ αk = 1
α 1 K uW A
n1
α K W
A w2 ≤ 2 u A
n2
A w1 ≤
k
with
I tot =
Σ
j=1
nj
n1 I j
More turns more resistance
more copper loss
23
Total power loss
Ptot = Pcu + Pfe
Power
loss
Co
ss P c
r lo
fe
ss P
u
P fe = K fe(ΔB) β A c l m
re l
o
ppe
Ptot = Pfe + Pcu
ρ(MLT )n 21 I 2tot
Pcu =
WAK u
VgDTs
B=
2n 1A c
Ptot
Co
There is a value of B
(or n1) that minimizes
the total power loss
Optimum ΔB
Prelab assignment for next week: use a
spreadsheet or other computer tool to compute
Ptot vs. n1, and find the optimum n1.
Then design your flyback transformer.
ECEN 4517
ΔB
24
Effect of transformer leakage inductance
Transformer model
Ll
ig
+ vl –
i
1:n
LM
Vg
+
D1
C
R
+
–
v
–
Q1
+
–
Voltage spike
caused by
leakage
inductance
di
vl = L l l
dt
{
• Ll induces a voltage spike across Q1
Vg + v/n
iRon
t
DTs
ECEN 4517
• Leakage inductance is effectively in
series with transistor Q1
• When MOSFET switches off, it
interrupts the current in Ll
vT(t)
vT(t)
• Leakage inductance Ll is caused by
imperfect coupling of primary and
secondary windings
25
If the peak magnitude of the
voltage spike exceeds the
voltage rating of the MOSFET,
then the MOSFET will fail.
Protection of Q1
using a voltage-clamp snubber
Snubber
{
Flyback transformer
ig
–
Rs
Vg
+
–
vs
1:n
Cs
+
D1
C
R
v
+
–
Q1
+
vT(t)
• Snubber provides a place
for current in leakage
inductance to flow after
Q1 has turned off
• Peak transistor voltage is
clamped to Vg + vs
• vs > V/n
–
• Energy stored in leakage
inductance (plus more) is
transferred to capacitor
Cs, then dissipated in Rs
Usually, Cs is large
Decreasing Rs decreases the peak transistor voltage but increases the
snubber power loss
See supplementary flyback notes for an example of estimating Cs and Rs
ECEN 4517
26
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